How to create a String with format? - ios

I need to create a String with format which can convert Int, Int64, Double, etc types into String. Using Objective-C, I can do it by:
NSString *str = [NSString stringWithFormat:#"%d , %f, %ld, %#", INT_VALUE, FLOAT_VALUE, DOUBLE_VALUE, STRING_VALUE];
How to do same but in Swift?

I think this could help you:
import Foundation
let timeNow = time(nil)
let aStr = String(format: "%#%x", "timeNow in hex: ", timeNow)
print(aStr)
Example result:
timeNow in hex: 5cdc9c8d

nothing special
let str = NSString(format:"%d , %f, %ld, %#", INT_VALUE, FLOAT_VALUE, LONG_VALUE, STRING_VALUE)

let str = "\(INT_VALUE), \(FLOAT_VALUE), \(DOUBLE_VALUE), \(STRING_VALUE)"
Update: I wrote this answer before Swift had String(format:) added to it's API. Use the method given by the top answer.

No NSString required!
String(format: "Value: %3.2f\tResult: %3.2f", arguments: [2.7, 99.8])
or
String(format:"Value: %3.2f\tResult: %3.2f", 2.7, 99.8)

I would argue that both
let str = String(format:"%d, %f, %ld", INT_VALUE, FLOAT_VALUE, DOUBLE_VALUE)
and
let str = "\(INT_VALUE), \(FLOAT_VALUE), \(DOUBLE_VALUE)"
are both acceptable since the user asked about formatting and both cases fit what they are asking for:
I need to create a string with format which can convert int, long, double etc. types into string.
Obviously the former allows finer control over the formatting than the latter, but that does not mean the latter is not an acceptable answer.

First read Official documentation for Swift language.
Answer should be
var str = "\(INT_VALUE) , \(FLOAT_VALUE) , \(DOUBLE_VALUE), \(STRING_VALUE)"
println(str)
Here
1) Any floating point value by default double
EX.
var myVal = 5.2 // its double by default;
-> If you want to display floating point value then you need to explicitly define such like a
EX.
var myVal:Float = 5.2 // now its float value;
This is far more clear.

let INT_VALUE=80
let FLOAT_VALUE:Double= 80.9999
let doubleValue=65.0
let DOUBLE_VALUE:Double= 65.56
let STRING_VALUE="Hello"
let str = NSString(format:"%d , %f, %ld, %#", INT_VALUE, FLOAT_VALUE, DOUBLE_VALUE, STRING_VALUE);
println(str);

The accepted answer is definitely the best general solution for this (i.e., just use the String(format:_:) method from Foundation) but...
If you are running Swift ≥ 5, you can leverage the new StringInterpolationProtocol protocol to give yourself some very nice syntax sugar for common string formatting use cases in your app.
Here is how the official documentation summarizes this new protocol:
Represents the contents of a string literal with interpolations while it’s being built up.
Some quick examples:
extension String.StringInterpolation {
/// Quick formatting for *floating point* values.
mutating func appendInterpolation(float: Double, decimals: UInt = 2) {
let floatDescription = String(format: "%.\(decimals)f%", float)
appendLiteral(floatDescription)
}
/// Quick formatting for *hexadecimal* values.
mutating func appendInterpolation(hex: Int) {
let hexDescription = String(format: "0x%X", hex)
appendLiteral(hexDescription)
}
/// Quick formatting for *percents*.
mutating func appendInterpolation(percent: Double, decimals: UInt = 2) {
let percentDescription = String(format: "%.\(decimals)f%%", percent * 100)
appendLiteral(percentDescription)
}
/// Formats the *elapsed time* since the specified start time.
mutating func appendInterpolation(timeSince startTime: TimeInterval, decimals: UInt = 2) {
let elapsedTime = CACurrentMediaTime() - startTime
let elapsedTimeDescription = String(format: "%.\(decimals)fs", elapsedTime)
appendLiteral(elapsedTimeDescription)
}
}
which could be used as:
let number = 1.2345
"Float: \(float: number)" // "Float: 1.23"
"Float: \(float: number, decimals: 1)" // "Float: 1.2"
let integer = 255
"Hex: \(hex: integer)" // "Hex: 0xFF"
let rate = 0.15
"Percent: \(percent: rate)" // "Percent: 15.00%"
"Percent: \(percent: rate, decimals: 0)" // "Percent: 15%"
let startTime = CACurrentMediaTime()
Thread.sleep(forTimeInterval: 2.8)
"∆t was \(timeSince: startTime)" // "∆t was 2.80s"
"∆t was \(timeSince: startTime, decimals: 0)" // "∆t was 3s"
This was introduced by SE-0228, so please be sure to read the original proposal for a deeper understanding of this new feature. Finally, the protocol documentation is helpful as well.

I know a lot's of time has passed since this publish, but I've fallen in a similar situation and create a simples class to simplify my life.
public struct StringMaskFormatter {
public var pattern : String = ""
public var replecementChar : Character = "*"
public var allowNumbers : Bool = true
public var allowText : Bool = false
public init(pattern:String, replecementChar:Character="*", allowNumbers:Bool=true, allowText:Bool=true)
{
self.pattern = pattern
self.replecementChar = replecementChar
self.allowNumbers = allowNumbers
self.allowText = allowText
}
private func prepareString(string:String) -> String {
var charSet : NSCharacterSet!
if allowText && allowNumbers {
charSet = NSCharacterSet.alphanumericCharacterSet().invertedSet
}
else if allowText {
charSet = NSCharacterSet.letterCharacterSet().invertedSet
}
else if allowNumbers {
charSet = NSCharacterSet.decimalDigitCharacterSet().invertedSet
}
let result = string.componentsSeparatedByCharactersInSet(charSet)
return result.joinWithSeparator("")
}
public func createFormattedStringFrom(text:String) -> String
{
var resultString = ""
if text.characters.count > 0 && pattern.characters.count > 0
{
var finalText = ""
var stop = false
let tempString = prepareString(text)
var formatIndex = pattern.startIndex
var tempIndex = tempString.startIndex
while !stop
{
let formattingPatternRange = formatIndex ..< formatIndex.advancedBy(1)
if pattern.substringWithRange(formattingPatternRange) != String(replecementChar) {
finalText = finalText.stringByAppendingString(pattern.substringWithRange(formattingPatternRange))
}
else if tempString.characters.count > 0 {
let pureStringRange = tempIndex ..< tempIndex.advancedBy(1)
finalText = finalText.stringByAppendingString(tempString.substringWithRange(pureStringRange))
tempIndex = tempIndex.advancedBy(1)
}
formatIndex = formatIndex.advancedBy(1)
if formatIndex >= pattern.endIndex || tempIndex >= tempString.endIndex {
stop = true
}
resultString = finalText
}
}
return resultString
}
}
The follow link send to the complete source code:
https://gist.github.com/dedeexe/d9a43894081317e7c418b96d1d081b25
This solution was base on this article:
http://vojtastavik.com/2015/03/29/real-time-formatting-in-uitextfield-swift-basics/

There is a simple solution I learned with "We <3 Swift" if you can't either import Foundation, use round() and/or does not want a String:
var number = 31.726354765
var intNumber = Int(number * 1000.0)
var roundedNumber = Double(intNumber) / 1000.0
result: 31.726

Use this following code:
let intVal=56
let floatval:Double=56.897898
let doubleValue=89.0
let explicitDaouble:Double=89.56
let stringValue:"Hello"
let stringValue="String:\(stringValue) Integer:\(intVal) Float:\(floatval) Double:\(doubleValue) ExplicitDouble:\(explicitDaouble) "

The beauty of String(format:) is that you can save a formatting string and then reuse it later in dozen of places. It also can be localized in this single place. Where as in case of the interpolation approach you must write it again and again.

Simple functionality is not included in Swift, expected because it's included in other languages, can often be quickly coded for reuse. Pro tip for programmers to create a bag of tricks file that contains all this reuse code.
So from my bag of tricks we first need string multiplication for use in indentation.
#inlinable func * (string: String, scalar: Int) -> String {
let array = [String](repeating: string, count: scalar)
return array.joined(separator: "")
}
and then the code to add commas.
extension Int {
#inlinable var withCommas:String {
var i = self
var retValue:[String] = []
while i >= 1000 {
retValue.append(String(format:"%03d",i%1000))
i /= 1000
}
retValue.append("\(i)")
return retValue.reversed().joined(separator: ",")
}
#inlinable func withCommas(_ count:Int = 0) -> String {
let retValue = self.withCommas
let indentation = count - retValue.count
let indent:String = indentation >= 0 ? " " * indentation : ""
return indent + retValue
}
}
I just wrote this last function so I could get the columns to line up.
The #inlinable is great because it takes small functions and reduces their functionality so they run faster.
You can use either the variable version or, to get a fixed column, use the function version. Lengths set less than the needed columns will just expand the field.
Now you have something that is pure Swift and does not rely on some old objective C routine for NSString.

Since String(format: "%s" ...) is crashing at run time, here is code to allow write something like "hello".center(42); "world".alignLeft(42):
extension String {
// note: symbol names match to nim std/strutils lib:
func align (_ boxsz: UInt) -> String {
self.withCString { String(format: "%\(boxsz)s", $0) }
}
func alignLeft (_ boxsz: UInt) -> String {
self.withCString { String(format: "%-\(boxsz)s", $0) }
}
func center (_ boxsz: UInt) -> String {
let n = self.count
guard boxsz > n else { return self }
let padding = boxsz - UInt(n)
let R = padding / 2
guard R > 0 else { return " " + self }
let L = (padding%2 == 0) ? R : (R+1)
return " ".withCString { String(format: "%\(L)s\(self)%\(R)s", $0,$0) }
}
}

Success to try it:
var letters:NSString = "abcdefghijkl"
var strRendom = NSMutableString.stringWithCapacity(strlength)
for var i=0; i<strlength; i++ {
let rndString = Int(arc4random() % 12)
//let strlk = NSString(format: <#NSString#>, <#CVarArg[]#>)
let strlk = NSString(format: "%c", letters.characterAtIndex(rndString))
strRendom.appendString(String(strlk))
}

Related

How can I change a UILabel into a number programmatically? [duplicate]

I'm trying to work out how to cast an Int into a String in Swift.
I figure out a workaround, using NSNumber but I'd love to figure out how to do it all in Swift.
let x : Int = 45
let xNSNumber = x as NSNumber
let xString : String = xNSNumber.stringValue
Converting Int to String:
let x : Int = 42
var myString = String(x)
And the other way around - converting String to Int:
let myString : String = "42"
let x: Int? = myString.toInt()
if (x != nil) {
// Successfully converted String to Int
}
Or if you're using Swift 2 or 3:
let x: Int? = Int(myString)
Check the Below Answer:
let x : Int = 45
var stringValue = "\(x)"
print(stringValue)
Here are 4 methods:
var x = 34
var s = String(x)
var ss = "\(x)"
var sss = toString(x)
var ssss = x.description
I can imagine that some people will have an issue with ss. But if you were looking to build a string containing other content then why not.
In Swift 3.0:
var value: Int = 10
var string = String(describing: value)
Swift 4:
let x:Int = 45
let str:String = String(describing: x)
Developer.Apple.com > String > init(describing:)
The String(describing:) initializer is the preferred way to convert an instance of any type to a string.
Custom String Convertible
Just for completeness, you can also use:
let x = 10.description
or any other value that supports a description.
Swift 4:
Trying to show the value in label without Optional() word.
here x is a Int value using.
let str:String = String(x ?? 0)
To save yourself time and hassle in the future you can make an Int extension. Typically I create a shared code file where I put extensions, enums, and other fun stuff. Here is what the extension code looks like:
extension Int
{
func toString() -> String
{
var myString = String(self)
return myString
}
}
Then later when you want to convert an int to a string you can just do something like:
var myNumber = 0
var myNumberAsString = myNumber.toString()
in swift 3.0 this is how we can convert Int to String and String to Int
//convert Integer to String in Swift 3.0
let theIntegerValue :Int = 123 // this can be var also
let theStringValue :String = String(theIntegerValue)
//convert String to Integere in Swift 3.0
let stringValue : String = "123"
let integerValue : Int = Int(stringValue)!
for whatever reason the accepted answer did not work for me. I went with this approach:
var myInt:Int = 10
var myString:String = toString(myInt)
Multiple ways to do this :
var str1:String="\(23)"
var str2:String=String(format:"%d",234)
let intAsString = 45.description // "45"
let stringAsInt = Int("45") // 45
Swift 2:
var num1 = 4
var numString = "56"
var sum2 = String(num1) + numString
var sum3 = Int(numString)
Swift String performance
A little bit about performance
UI Testing Bundle on iPhone 7(real device) with iOS 14
let i = 0
lt result1 = String(i) //0.56s 5890kB
lt result2 = "\(i)" //0.624s 5900kB
lt result3 = i.description //0.758s 5890kB
import XCTest
class ConvertIntToStringTests: XCTestCase {
let count = 1_000_000
func measureFunction(_ block: () -> Void) {
let metrics: [XCTMetric] = [
XCTClockMetric(),
XCTMemoryMetric()
]
let measureOptions = XCTMeasureOptions.default
measureOptions.iterationCount = 5
measure(metrics: metrics, options: measureOptions) {
block()
}
}
func testIntToStringConstructor() {
var result = ""
measureFunction {
for i in 0...count {
result += String(i)
}
}
}
func testIntToStringInterpolation() {
var result = ""
measureFunction {
for i in 0...count {
result += "\(i)"
}
}
}
func testIntToStringDescription() {
var result = ""
measureFunction {
for i in 0...count {
result += i.description
}
}
}
}
iam using this simple approach
String to Int:
var a = Int()
var string1 = String("1")
a = string1.toInt()
and from Int to String:
var a = Int()
a = 1
var string1 = String()
string1= "\(a)"
Convert Unicode Int to String
For those who want to convert an Int to a Unicode string, you can do the following:
let myInteger: Int = 97
// convert Int to a valid UnicodeScalar
guard let myUnicodeScalar = UnicodeScalar(myInteger) else {
return ""
}
// convert UnicodeScalar to String
let myString = String(myUnicodeScalar)
// results
print(myString) // a
Or alternatively:
let myInteger: Int = 97
if let myUnicodeScalar = UnicodeScalar(myInteger) {
let myString = String(myUnicodeScalar)
}
I prefer using String Interpolation
let x = 45
let string = "\(x)"
Each object has some string representation. This makes things simpler. For example if you need to create some String with multiple values. You can also do any math in it or use some conditions
let text = "\(count) \(count > 1 ? "items" : "item") in the cart. Sum: $\(sum + shippingPrice)"
exampleLabel.text = String(yourInt)
To convert String into Int
var numberA = Int("10")
Print(numberA) // It will print 10
To covert Int into String
var numberA = 10
1st way)
print("numberA is \(numberA)") // It will print 10
2nd way)
var strSomeNumber = String(numberA)
or
var strSomeNumber = "\(numberA)"
let a =123456888
var str = String(a)
OR
var str = a as! String
In swift 3.0, you may change integer to string as given below
let a:String = String(stringInterpolationSegment: 15)
Another way is
let number: Int = 15
let _numberInStringFormate: String = String(number)
//or any integer number in place of 15
If you like swift extension, you can add following code
extension Int
{
var string:String {
get {
return String(self)
}
}
}
then, you can get string by the method you just added
var x = 1234
var s = x.string
let Str = "12"
let num: Int = 0
num = Int (str)

Strange return value after generating random string (Swift 3)

I've created a method which generates and returns a random string of both letters and numbers, but for some reason I only get a string with numbers and the length of the string doesn't come close to what I asked it to be. A few examples of strings that have been returned: "478388299949939566" (inserted 18 as the length), "3772919388584334" (inserted 9 as the length), "2293010089409293945" (inserted 6 as the length). Anyone can see what I'm missing here?
func generateRandomStringWithLength(length:Int) -> String {
let randomString:NSMutableString = NSMutableString(capacity: length)
let letters:NSMutableString = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
for index in 0...length {
let randomIndex:Int = Int(arc4random_uniform(UInt32(62)))
randomString.append("\(letters.character(at: randomIndex))")
}
return String(randomString)
}
Your problem is here:
letters.character(at: randomIndex)
it's function returns the character at a given UTF-16 code unit index, not not just a character
Here is my version, I guess its more swiftly.
func generateRandomStringWithLength(length: Int) -> String {
var randomString = ""
let letters = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
for _ in 1...length {
let randomIndex = Int(arc4random_uniform(UInt32(letters.characters.count)))
let a = letters.index(letters.startIndex, offsetBy: randomIndex)
randomString += String(letters[a])
}
return randomString
}
generateRandomStringWithLength(length: 5)
Use this:
func generateRandomStringWithLength(length:Int) -> String {
let randomString:NSMutableString = NSMutableString(capacity: length)
let letters:NSMutableString = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
var i: Int = 0
while i < length {
let randomIndex:Int = Int(arc4random_uniform(UInt32(letters.length)))
randomString.appendString("\(Character( UnicodeScalar( letters.characterAtIndex(randomIndex))))")
i += 1
}
return String(randomString)
}
Calling generateRandomStringWithLength method:
print(generateRandomStringWithLength(5))
print(generateRandomStringWithLength(10))
print(generateRandomStringWithLength(20))
print(generateRandomStringWithLength(7))
print(generateRandomStringWithLength(14))
Sample Output:
GIrqb
nWmieQRVdk
r0It9V1xkGFRa2HVwtCw
RLIRuVI
nXnFGV2LQ3CjbD

Swift Anagram checker

I am attempting to build an anagram checker for swift. This is my code. In case you don't know an anagram checker checks if two strings have the same characters in them but, order does not matter.
func checkForAnagram(#firstString: String, #secondString: String) -> Bool {
var firstStringArray: [Character] = []
var secondStringArray: [Character] = []
/* if case matters delete the next four lines
and make sure your variables are not constants */
var first = firstString
var second = secondString
first = first.lowercaseString
second = second.lowercaseString
for charactersOne in first {
firstStringArray += [charactersOne]
}
for charactersTwo in second {
secondStringArray += [charactersTwo]
}
if firstStringArray.count != secondStringArray.count {
return false
} else {
for elements in firstStringArray {
if secondStringArray.contains(elements){
return true
} else {
return false
}
}
}
}
var a = "Hello"
var b = "oellh"
var c = "World"
checkForAnagram(firstString: a, secondString: b)
I am getting an error message of.
'[Character]' does not have a member 'contains'
The accepted answer is compact and elegant, but very inefficient if compared to other solutions.
I'll now propose and discuss the implementation of a few variants of anagram checker. To measure performance, I'll use the different variants to find the anagrams of a given word out of an array of 50,000+ words.
// Variant 1: Sorting of Character
// Measured time: 30.46 s
func anagramCheck1(a: String, b: String) -> Bool {
return a.characters.sorted() == b.characters.sorted()
}
This is essentially the solution of the accepted answer, written in Swift 3 syntax. It's very slow because Swift's String, unlike NSString, is based on Character, which handles Unicode characters properly.
A more efficient solution exploits the NSCountedSet class, which allows us to represent a string as a set of characters, each with its own count. Two strings are anagrams if they map to the same NSCountedSet.
Note: checking string lengths as a precondition makes the implementation always more efficient.
// Variant 2: NSCountedSet of Character
// Measured time: 4.81 s
func anagramCheck2(a: String, b: String) -> Bool {
guard a.characters.count == b.characters.count else { return false }
let aSet = NSCountedSet()
let bSet = NSCountedSet()
for c in a.characters {
aSet.add(c)
}
for c in b.characters {
bSet.add(c)
}
return aSet == bSet
}
Better but not excellent. Here, one of the "culprits" is the use of the native Swift Character type (from Swift's String). Moving back to good old Objective-C types (NSString and unichar) makes things more efficient.
// Variant 3: NSCountedSet of unichar
// Measured time: 1.31 s
func anagramCheck3(a: String, b: String) -> Bool {
let aString = a as NSString
let bString = b as NSString
let length = aString.length
guard length == bString.length else { return false }
let aSet = NSCountedSet()
let bSet = NSCountedSet()
for i in 0..<length {
aSet.add(aString.character(at: i))
bSet.add(bString.character(at: i))
}
return aSet == bSet
}
Using NSCountedSet is fine, but before we compare two NSCountedSet objects, we fully populate them. A useful alternative is to fully populate the NSCountedSet for only one of the two strings, and then, while we populate the NSCountedSet for the other string, we fail early if the other string contains a character that is not found in the NSCountedSet of the first string.
// Variant 4: NSCountedSet of unichar and early exit
// Measured time: 1.07 s
func anagramCheck4(a: String, b: String) -> Bool {
let aString = a as NSString
let bString = b as NSString
let length = aString.length
guard length == bString.length else { return false }
let aSet = NSCountedSet()
let bSet = NSCountedSet()
for i in 0..<length {
aSet.add(aString.character(at: i))
}
for i in 0..<length {
let c = bString.character(at: i)
if bSet.count(for: c) >= aSet.count(for: c) {
return false
}
bSet.add(c)
}
return true
}
This is about the best timing we are going to get (with Swift). However, for completeness, let me discuss one more variant of this kind.
The next alternative exploits a Swift Dictionary of type [unichar: Int] to store the number of repetitions for each character instead of NSCountedSet. It's slightly slower than the previous two variants, but we can reuse it later to obtain a faster implementation.
// Variant 5: counting repetitions with [unichar:Int]
// Measured time: 1.36
func anagramCheck5(a: String, b: String) -> Bool {
let aString = a as NSString
let bString = b as NSString
let length = aString.length
guard length == bString.length else { return false }
var aDic = [unichar:Int]()
var bDic = [unichar:Int]()
for i in 0..<length {
let c = aString.character(at: i)
aDic[c] = (aDic[c] ?? 0) + 1
}
for i in 0..<length {
let c = bString.character(at: i)
let count = (bDic[c] ?? 0) + 1
if count > aDic[c] ?? 0 {
return false
}
bDic[c] = count
}
return true
}
Note that a vanilla Objective-C implementation using NSCountedSet, corresponding to Variant 3, is faster than all the previous versions by a rather large margin.
// Variant 6: Objective-C and NSCountedSet
// Measured time: 0.65 s
- (BOOL)anagramChecker:(NSString *)a with:(NSString *)b {
if (a.length != b.length) {
return NO;
}
NSCountedSet *aSet = [[NSCountedSet alloc] init];
NSCountedSet *bSet = [[NSCountedSet alloc] init];
for (int i = 0; i < a.length; i++) {
[aSet addObject:#([a characterAtIndex:i])];
[bSet addObject:#([b characterAtIndex:i])];
}
return [aSet isEqual:bSet];
}
Another way we can improve upon the previous attempts is to observe that, if we need to find the anagram of a given word, we might as well consider that word as fixed, and thus we could build the corresponding structure (NSCountedSet, Dictionary, ...) for that word only once.
// Finding all the anagrams of word in words
// Variant 7: counting repetitions with [unichar:Int]
// Measured time: 0.58 s
func anagrams(word: String, from words: [String]) -> [String] {
let anagrammedWord = word as NSString
let length = anagrammedWord.length
var aDic = [unichar:Int]()
for i in 0..<length {
let c = anagrammedWord.character(at: i)
aDic[c] = (aDic[c] ?? 0) + 1
}
let foundWords = words.filter {
let string = $0 as NSString
guard length == string.length else { return false }
var bDic = [unichar:Int]()
for i in 0..<length {
let c = string.character(at: i)
let count = (bDic[c] ?? 0) + 1
if count > aDic[c] ?? 0 {
return false
}
bDic[c] = count
}
return true
}
return foundWords
}
Now, in the previous variant we have counted with a [unichar:Int] Dictionary. This proves slightly more efficient than using an NSCountedSet of unichar, either with early exit (0.60 s) or without (0.87 s).
You should try
func checkForAnagram(firstString firstString: String, secondString: String) -> Bool {
return firstString.lowercaseString.characters.sort() == secondString.lowercaseString.characters.sort()
}
func checkAnagrams(str1: String, str2: String) -> Bool {
guard str1.count == str2.count else { return false }
var dictionary = Dictionary<Character, Int>()
for index in 0..<str1.count {
let value1 = str1[str1.index(str1.startIndex, offsetBy: index)]
let value2 = str2[str2.index(str2.startIndex, offsetBy: index)]
dictionary[value1] = (dictionary[value1] ?? 0) + 1
dictionary[value2] = (dictionary[value2] ?? 0) - 1
}
return !dictionary.contains(where: {(_, value) in
return value != 0
})
}
Time complexity - O(n)
// Make sure name your variables correctly so you won't confuse
// Mutate the constants parameter, lowercase to handle capital letters and the sorted them to compare both. Finally check is there are equal return true or false.
func anagram(str1: String, srt2: String)->Bool{
let string1 = str1.lowercased().sorted()
let string2 = srt2.lowercased().sorted()
if string1 == string2 {
return true
}
return false
}
// This answer also would work
// Convert your parameters on Array, then sorted them and compare them
func ana(str1: String, str2: String)->Bool{
let a = Array(str1)
let b = Array(str2)
if a.sorted() == b.sorted() {
return true
}
return false
}
Don't forget whitespaces
func isAnagram(_ stringOne: String, stringTwo: String) -> Bool {
return stringOne.lowercased().sorted().filter { $0 != " "} stringTwo.lowercased().sorted().filter { $0 != " "}
}
Swift 4.1 Function will give you 3 questions answer for Anagram :-
1. Input Strings (a,b) are Anagram ? //Bool
2. If not an Anagram then number of count require to change Characters in strings(a,b) to make them anagram ? // Int
3. If not an Anagram then list of Characters needs to be change in strings(a,b) to make them anagram ? // [Character]
STEP 1:- Copy and Paste below function in to your required class:-
//MARK:- Anagram checker
func anagramChecker(a:String,b:String) -> (Bool,Int,[Character]) {
var aCharacters = Array(a)
var bCharacters = Array(b)
var count = 0
var isAnagram = true
var replacementRequiredWords:[Character] = [Character]()
if aCharacters.count == bCharacters.count {
let listA = aCharacters.filter { !bCharacters.contains($0) }
for i in 0 ..< listA.count {
if !replacementRequiredWords.contains(listA[i]) {
count = count + 1
replacementRequiredWords.append(listA[i])
isAnagram = false
}
}
let listB = bCharacters.filter { !aCharacters.contains($0) }
for i in 0 ..< listB.count {
if !replacementRequiredWords.contains(listB[i]) {
count = count + 1
replacementRequiredWords.append(listB[i])
isAnagram = false
}
}
}else{
//cant be an anagram
count = -1
}
return (isAnagram,count,replacementRequiredWords)
}
STEP 2 :- Make two Input Strings for test
// Input Strings
var a = "aeb"
var b = "abs"
STEP 3:- Print results :-
print("isAnagram : \(isAnagram(a: a, b: b).0)")
print("number of count require to change strings in anagram : \(isAnagram(a: a, b: b).1)")//-1 will come in case of cant be a Anagram
print("list of Characters needs to be change : \(isAnagram(a: a, b: b).2)")
Results of above exercise:-
isAnagram : false
number of count require to change strings in anagram : 2
list of Characters needs to be change : ["e", "s"]
Hope this 10 minutes exercise will give some support to my Swift
family for solving Anagram related problems easily. :)
We can use dictionary to construct a new data structure container. Then compare the value by key/character of the string.
func anagram(str1: String, str2 : String) -> Bool {
var dict1 = [Character: Int]()
var dict2 = [Character:Int]()
for i in str1 {
if let count = dict1[i] {
dict1[i] = count + 1
} else {
dict1[i] = 1
}
}
for j in str2 {
if let count = dict2[j] {
dict2[j] = count + 1
} else {
dict2[j] = 1
}
}
return dict1 == dict2 ? true : false
}
// input -> "anna", "aann"
// The count will look like:
// ["a": 2, "n": 2] & ["a": 2, "n": 2]
// then return true
Another easy that I just realise doing an Anagram function in Swift 5.X
func checkForAnagram(firstString firstString: String, secondString: String) -> Bool {
return !firstString.isEmpty && firstString.sorted() == secondString.sorted()
}
class Solution {
func isAnagram(_ s: String, _ t: String) -> Bool {
guard s.count == t.count else { return false }
let dictS = s.reduce(into: [Character: Int]()) { $0[$1, default: 0] += 1 }
let dictT = t.reduce(into: [Character: Int]()) { $0[$1, default: 0] += 1 }
for letter in s {
if let count = dictS[letter] {
guard count == dictT[letter] else { return false }
}
}
return true
}
}
Check two strings are anagram using inout method in Swift
func checkAnagramString(str1: inout String, str2: inout String)-> Bool{
var result:Bool = false
str1 = str1.lowercased().trimmingCharacters(in: .whitespace)
str2 = str2.lowercased().trimmingCharacters(in: .whitespaces)
if (str1.count != str2.count) {
return result
}
for c in str1 {
if str2.contains(c){
result = true
}
else{
result = false
return result
}
}
return result
}
Call function to check strings are anagram or not
var str1 = "tommarvoloriddle"
var str2 = "iamlordvoldemort"
print(checkAnagramString(str1: &str1, str2: &str2)) //Output = true.
func isAnagram(word1: String, word2: String) -> Bool {
let set1 = Set(word1)
let set2 = Set(word2)
return set1 == set2
}
or
func isAnagram(word1: String,word2: String) -> Bool {
return word1.lowercased().sorted() == word2.lowercased().sorted()
}

Swift - How to remove a decimal from a float if the decimal is equal to 0?

I'm displaying a distance with one decimal, and I would like to remove this decimal in case it is equal to 0 (ex: 1200.0Km), how could I do that in swift?
I'm displaying this number like this:
let distanceFloat: Float = (currentUser.distance! as NSString).floatValue
distanceLabel.text = String(format: "%.1f", distanceFloat) + "Km"
Swift 3/4:
var distanceFloat1: Float = 5.0
var distanceFloat2: Float = 5.540
var distanceFloat3: Float = 5.03
extension Float {
var clean: String {
return self.truncatingRemainder(dividingBy: 1) == 0 ? String(format: "%.0f", self) : String(self)
}
}
print("Value \(distanceFloat1.clean)") // 5
print("Value \(distanceFloat2.clean)") // 5.54
print("Value \(distanceFloat3.clean)") // 5.03
Swift 2 (Original answer)
let distanceFloat: Float = (currentUser.distance! as NSString).floatValue
distanceLabel.text = String(format: distanceFloat == floor(distanceFloat) ? “%.0f" : "%.1f", distanceFloat) + "Km"
Or as an extension:
extension Float {
var clean: String {
return self % 1 == 0 ? String(format: "%.0f", self) : String(self)
}
}
Use NSNumberFormatter:
let formatter = NumberFormatter()
formatter.minimumFractionDigits = 0
formatter.maximumFractionDigits = 2
// Avoid not getting a zero on numbers lower than 1
// Eg: .5, .67, etc...
formatter.numberStyle = .decimal
let nums = [3.0, 5.1, 7.21, 9.311, 600.0, 0.5677, 0.6988]
for num in nums {
print(formatter.string(from: num as NSNumber) ?? "n/a")
}
Returns:
3
5.1
7.21
9.31
600
0.57
0.7
extension is the powerful way to do it.
Extension:
Code for Swift 2 (not Swift 3 or newer):
extension Float {
var cleanValue: String {
return self % 1 == 0 ? String(format: "%.0f", self) : String(self)
}
}
Usage:
var sampleValue: Float = 3.234
print(sampleValue.cleanValue)
3.234
sampleValue = 3.0
print(sampleValue.cleanValue)
3
sampleValue = 3
print(sampleValue.cleanValue)
3
Sample Playground file is here.
Update of accepted answer for swift 3:
extension Float {
var cleanValue: String {
return self.truncatingRemainder(dividingBy: 1) == 0 ? String(format: "%.0f", self) : String(self)
}
}
usage would just be:
let someValue: Float = 3.0
print(someValue.cleanValue) //prints 3
To format it to String, follow this pattern
let aFloat: Float = 1.123
let aString: String = String(format: "%.0f", aFloat) // "1"
let aString: String = String(format: "%.1f", aFloat) // "1.1"
let aString: String = String(format: "%.2f", aFloat) // "1.12"
let aString: String = String(format: "%.3f", aFloat) // "1.123"
To cast it to Int, follow this pattern
let aInt: Int = Int(aFloat) // "1"
When you use String(format: initializer, Swift will automatically round the final digit as needed based on the following number.
You can use an extension as already mentioned, this solution is a little shorter though:
extension Float {
var shortValue: String {
return String(format: "%g", self)
}
}
Example usage:
var sample: Float = 3.234
print(sample.shortValue)
Swift 5
for Double it's same as #Frankie's answer for float
var dec: Double = 1.0
dec.clean // 1
for the extension
extension Double {
var clean: String {
return self.truncatingRemainder(dividingBy: 1) == 0 ? String(format: "%.0f", self) : String(self)
}
}
Swift 5.5 makes it easy
Just use the new formatted() api with a default FloatingPointFormatStyle:
let values: [Double] = [1.0, 4.5, 100.0, 7]
for value in values {
print(value.formatted(FloatingPointFormatStyle()))
}
// prints "1, 4.5, 100, 7"
In Swift 4 try this.
extension CGFloat{
var cleanValue: String{
//return String(format: 1 == floor(self) ? "%.0f" : "%.2f", self)
return self.truncatingRemainder(dividingBy: 1) == 0 ? String(format: "%.0f", self) : String(format: "%.2f", self)//
}
}
//How to use - if you enter more then two-character after (.)point, it's automatically cropping the last character and only display two characters after the point.
let strValue = "32.12"
print(\(CGFloat(strValue).cleanValue)
Formatting with maximum fraction digits, without trailing zeros
This scenario is good when a custom output precision is desired.
This solution seems roughly as fast as NumberFormatter + NSNumber solution from MirekE, but one benefit could be that we're avoiding NSObject here.
extension Double {
func string(maximumFractionDigits: Int = 2) -> String {
let s = String(format: "%.\(maximumFractionDigits)f", self)
var offset = -maximumFractionDigits - 1
for i in stride(from: 0, to: -maximumFractionDigits, by: -1) {
if s[s.index(s.endIndex, offsetBy: i - 1)] != "0" {
offset = i
break
}
}
return String(s[..<s.index(s.endIndex, offsetBy: offset)])
}
}
(works also with extension Float, but not the macOS-only type Float80)
Usage: myNumericValue.string(maximumFractionDigits: 2) or myNumericValue.string()
Output for maximumFractionDigits: 2:
1.0 → "1"
0.12 → "0.12"
0.012 → "0.01"
0.0012 → "0"
0.00012 → "0"
Simple :
Int(floor(myFloatValue))
NSNumberFormatter is your friend
let distanceFloat: Float = (currentUser.distance! as NSString).floatValue
let numberFormatter = NSNumberFormatter()
numberFormatter.positiveFormat = "###0.##"
let distance = numberFormatter.stringFromNumber(NSNumber(float: distanceFloat))!
distanceLabel.text = distance + " Km"
Here's the full code.
let numberA: Float = 123.456
let numberB: Float = 789.000
func displayNumber(number: Float) {
if number - Float(Int(number)) == 0 {
println("\(Int(number))")
} else {
println("\(number)")
}
}
displayNumber(numberA) // console output: 123.456
displayNumber(numberB) // console output: 789
Here's the most important line in-depth.
func displayNumber(number: Float) {
Strips the float's decimal digits with Int(number).
Returns the stripped number back to float to do an operation with Float(Int(number)).
Gets the decimal-digit value with number - Float(Int(number))
Checks the decimal-digit value is empty with if number - Float(Int(number)) == 0
The contents within the if and else statements doesn't need explaining.
This might be helpful too.
extension Float {
func cleanValue() -> String {
let intValue = Int(self)
if self == 0 {return "0"}
if self / Float (intValue) == 1 { return "\(intValue)" }
return "\(self)"
}
}
Usage:
let number:Float = 45.23230000
number.cleanValue()
Maybe stringByReplacingOccurrencesOfString could help you :)
let aFloat: Float = 1.000
let aString: String = String(format: "%.1f", aFloat) // "1.0"
let wantedString: String = aString.stringByReplacingOccurrencesOfString(".0", withString: "") // "1"

Converting String to Int with Swift

The application basically calculates acceleration by inputting Initial and final velocity and time and then use a formula to calculate acceleration. However, since the values in the text boxes are string, I am unable to convert them to integers.
#IBOutlet var txtBox1 : UITextField
#IBOutlet var txtBox2 : UITextField
#IBOutlet var txtBox3 : UITextField
#IBOutlet var lblAnswer : UILabel
#IBAction func btn1(sender : AnyObject) {
let answer1 = "The acceleration is"
var answer2 = txtBox1
var answer3 = txtBox2
var answer4 = txtBox3
Updated answer for Swift 2.0+:
toInt() method gives an error, as it was removed from String in Swift 2.x. Instead, the Int type now has an initializer that accepts a String:
let a: Int? = Int(firstTextField.text)
let b: Int? = Int(secondTextField.text)
Basic Idea, note that this only works in Swift 1.x (check out ParaSara's answer to see how it works in Swift 2.x):
// toInt returns optional that's why we used a:Int?
let a:Int? = firstText.text.toInt() // firstText is UITextField
let b:Int? = secondText.text.toInt() // secondText is UITextField
// check a and b before unwrapping using !
if a && b {
var ans = a! + b!
answerLabel.text = "Answer is \(ans)" // answerLabel ie UILabel
} else {
answerLabel.text = "Input values are not numeric"
}
Update for Swift 4
...
let a:Int? = Int(firstText.text) // firstText is UITextField
let b:Int? = Int(secondText.text) // secondText is UITextField
...
myString.toInt() - convert the string value into int .
Swift 3.x
If you have an integer hiding inside a string, you can convertby using the integer's constructor, like this:
let myInt = Int(textField.text)
As with other data types (Float and Double) you can also convert by using NSString:
let myString = "556"
let myInt = (myString as NSString).integerValue
You can use NSNumberFormatter().numberFromString(yourNumberString). It's great because it returns an an optional that you can then test with if let to determine if the conversion was successful.
eg.
var myString = "\(10)"
if let myNumber = NSNumberFormatter().numberFromString(myString) {
var myInt = myNumber.integerValue
// do what you need to do with myInt
} else {
// what ever error code you need to write
}
Swift 5
var myString = "\(10)"
if let myNumber = NumberFormatter().number(from: myString) {
var myInt = myNumber.intValue
// do what you need to do with myInt
} else {
// what ever error code you need to write
}
edit/update: Xcode 11.4 • Swift 5.2
Please check the comments through the code
IntegerField.swift file contents:
import UIKit
class IntegerField: UITextField {
// returns the textfield contents, removes non digit characters and converts the result to an integer value
var value: Int { string.digits.integer ?? 0 }
var maxValue: Int = 999_999_999
private var lastValue: Int = 0
override func willMove(toSuperview newSuperview: UIView?) {
// adds a target to the textfield to monitor when the text changes
addTarget(self, action: #selector(editingChanged), for: .editingChanged)
// sets the keyboard type to digits only
keyboardType = .numberPad
// set the text alignment to right
textAlignment = .right
// sends an editingChanged action to force the textfield to be updated
sendActions(for: .editingChanged)
}
// deletes the last digit of the text field
override func deleteBackward() {
// note that the field text property default value is an empty string so force unwrap its value is safe
// note also that collection remove at requires a non empty collection which is true as well in this case so no need to check if the collection is not empty.
text!.remove(at: text!.index(before: text!.endIndex))
// sends an editingChanged action to force the textfield to be updated
sendActions(for: .editingChanged)
}
#objc func editingChanged() {
guard value <= maxValue else {
text = Formatter.decimal.string(for: lastValue)
return
}
// This will format the textfield respecting the user device locale and settings
text = Formatter.decimal.string(for: value)
print("Value:", value)
lastValue = value
}
}
You would need to add those extensions to your project as well:
Extensions UITextField.swift file contents:
import UIKit
extension UITextField {
var string: String { text ?? "" }
}
Extensions Formatter.swift file contents:
import Foundation
extension Formatter {
static let decimal = NumberFormatter(numberStyle: .decimal)
}
Extensions NumberFormatter.swift file contents:
import Foundation
extension NumberFormatter {
convenience init(numberStyle: Style) {
self.init()
self.numberStyle = numberStyle
}
}
Extensions StringProtocol.swift file contents:
extension StringProtocol where Self: RangeReplaceableCollection {
var digits: Self { filter(\.isWholeNumber) }
var integer: Int? { Int(self) }
}
Sample project
swift 4.0
let stringNumber = "123"
let number = Int(stringNumber) //here number is of type "Int?"
//using Forced Unwrapping
if number != nil {
//string is converted to Int
}
you could also use Optional Binding other than forced binding.
eg:
if let number = Int(stringNumber) {
// number is of type Int
}
In Swift 4.2 and Xcode 10.1
let string = "789"
if let intValue = Int(string) {
print(intValue)
}
let integerValue = 789
let stringValue = String(integerValue)
OR
let stringValue = "\(integerValue)"
print(stringValue)
//Xcode 8.1 and swift 3.0
We can also handle it by Optional Binding, Simply
let occur = "10"
if let occ = Int(occur) {
print("By optional binding :", occ*2) // 20
}
Swift 3
The simplest and more secure way is:
#IBOutlet var textFieldA : UITextField
#IBOutlet var textFieldB : UITextField
#IBOutlet var answerLabel : UILabel
#IBAction func calculate(sender : AnyObject) {
if let intValueA = Int(textFieldA),
let intValueB = Int(textFieldB) {
let result = intValueA + intValueB
answerLabel.text = "The acceleration is \(result)"
}
else {
answerLabel.text = "The value \(intValueA) and/or \(intValueB) are not a valid integer value"
}
}
Avoid invalid values setting keyboard type to number pad:
textFieldA.keyboardType = .numberPad
textFieldB.keyboardType = .numberPad
In Swift 4:
extension String {
var numberValue:NSNumber? {
let formatter = NumberFormatter()
formatter.numberStyle = .decimal
return formatter.number(from: self)
}
}
let someFloat = "12".numberValue
Useful for String to Int and other type
extension String {
//Converts String to Int
public func toInt() -> Int? {
if let num = NumberFormatter().number(from: self) {
return num.intValue
} else {
return nil
}
}
//Converts String to Double
public func toDouble() -> Double? {
if let num = NumberFormatter().number(from: self) {
return num.doubleValue
} else {
return nil
}
}
/// EZSE: Converts String to Float
public func toFloat() -> Float? {
if let num = NumberFormatter().number(from: self) {
return num.floatValue
} else {
return nil
}
}
//Converts String to Bool
public func toBool() -> Bool? {
return (self as NSString).boolValue
}
}
Use it like :
"123".toInt() // 123
i have made a simple program, where you have 2 txt field you take input form the user and add them to make it simpler to understand please find the code below.
#IBOutlet weak var result: UILabel!
#IBOutlet weak var one: UITextField!
#IBOutlet weak var two: UITextField!
#IBAction func add(sender: AnyObject) {
let count = Int(one.text!)
let cal = Int(two.text!)
let sum = count! + cal!
result.text = "Sum is \(sum)"
}
hope this helps.
Swift 3.0
Try this, you don't need to check for any condition I have done everything just use this function. Send anything string, number, float, double ,etc,. you get a number as a value or 0 if it is unable to convert your value
Function:
func getNumber(number: Any?) -> NSNumber {
guard let statusNumber:NSNumber = number as? NSNumber else
{
guard let statString:String = number as? String else
{
return 0
}
if let myInteger = Int(statString)
{
return NSNumber(value:myInteger)
}
else{
return 0
}
}
return statusNumber
}
Usage:
Add the above function in code and to convert use
let myNumber = getNumber(number: myString)
if the myString has a number or string it returns the number else it returns 0
Example 1:
let number:String = "9834"
print("printing number \(getNumber(number: number))")
Output: printing number 9834
Example 2:
let number:Double = 9834
print("printing number \(getNumber(number: number))")
Output: printing number 9834
Example 3:
let number = 9834
print("printing number \(getNumber(number: number))")
Output: printing number 9834
About int() and Swift 2.x: if you get a nil value after conversion check if you try to convert a string with a big number (for example: 1073741824), in this case try:
let bytesInternet : Int64 = Int64(bytesInternetString)!
Latest swift3 this code is simply to convert string to int
let myString = "556"
let myInt = Int(myString)
Because a string might contain non-numerical characters you should use a guard to protect the operation. Example:
guard let labelInt:Int = Int(labelString) else {
return
}
useLabelInt()
I recently got the same issue. Below solution is work for me:
let strValue = "123"
let result = (strValue as NSString).integerValue
Swift5 float or int string to int:
extension String {
func convertStringToInt() -> Int {
return Int(Double(self) ?? 0.0)
}
}
let doubleStr = "4.2"
// print 4
print(doubleStr.convertStringToInt())
let intStr = "4"
// print 4
print(intStr.convertStringToInt())
Use this:
// get the values from text boxes
let a:Double = firstText.text.bridgeToObjectiveC().doubleValue
let b:Double = secondText.text.bridgeToObjectiveC().doubleValue
// we checking against 0.0, because above function return 0.0 if it gets failed to convert
if (a != 0.0) && (b != 0.0) {
var ans = a + b
answerLabel.text = "Answer is \(ans)"
} else {
answerLabel.text = "Input values are not numberic"
}
OR
Make your UITextField KeyboardType as DecimalTab from your XIB or storyboard, and remove any if condition for doing any calculation, ie.
var ans = a + b
answerLabel.text = "Answer is \(ans)"
Because keyboard type is DecimalPad there is no chance to enter other 0-9 or .
Hope this help !!
// To convert user input (i.e string) to int for calculation.I did this , and it works.
let num:Int? = Int(firstTextField.text!);
let sum:Int = num!-2
print(sum);
This works for me
var a:Int? = Int(userInput.text!)
for Swift3.x
extension String {
func toInt(defaultValue: Int) -> Int {
if let n = Int(self.trimmingCharacters(in: CharacterSet.whitespacesAndNewlines)) {
return n
} else {
return defaultValue
}
}
}
Swift 4, Swift 5
There are different cases to convert from something to something data type, it depends the input.
If the input data type is Any, we have to use as before convert to actual data type, then convert to data type what we want. For example:
func justGetDummyString() -> Any {
return "2000"
}
let dummyString: String = (justGetDummyString() as? String) ?? "" // output = "2000"
let dummyInt: Int = Int(dummyString) ?? 0 // output = 2000
for Alternative solution. You can use extension a native type. You can test with playground.
extension String {
func add(a: Int) -> Int? {
if let b = Int(self) {
return b + a
}
else {
return nil
}
}
}
"2".add(1)
My solution is to have a general extension for string to int conversion.
extension String {
// default: it is a number suitable for your project if the string is not an integer
func toInt(default: Int) -> Int {
if let result = Int(self) {
return result
}
else {
return default
}
}
}
#IBAction func calculateAclr(_ sender: Any) {
if let addition = addition(arrayString: [txtBox1.text, txtBox2.text, txtBox3.text]) {
print("Answer = \(addition)")
lblAnswer.text = "\(addition)"
}
}
func addition(arrayString: [Any?]) -> Int? {
var answer:Int?
for arrayElement in arrayString {
if let stringValue = arrayElement, let intValue = Int(stringValue) {
answer = (answer ?? 0) + intValue
}
}
return answer
}
Question : string "4.0000" can not be convert into integer using Int("4.000")?
Answer : Int() check string is integer or not if yes then give you integer and otherwise nil. but Float or Double can convert any number string to respective Float or Double without giving nil. Example if you have "45" integer string but using Float("45") gives you 45.0 float value or using Double("4567") gives you 45.0.
Solution : NSString(string: "45.000").integerValue or Int(Float("45.000")!)! to get correct result.
An Int in Swift contains an initializer that accepts a String. It returns an optional Int? as the conversion can fail if the string contains not a number.
By using an if let statement you can validate whether the conversion succeeded.
So your code become something like this:
#IBOutlet var txtBox1 : UITextField
#IBOutlet var txtBox2 : UITextField
#IBOutlet var txtBox3 : UITextField
#IBOutlet var lblAnswer : UILabel
#IBAction func btn1(sender : AnyObject) {
let answer1 = "The acceleration is"
var answer2 = txtBox1
var answer3 = txtBox2
var answer4 = txtBox3
if let intAnswer = Int(txtBox1.text) {
// Correctly converted
}
}
Swift 5.0 and Above
Working
In case if you are splitting the String it creates two substrings and not two Strings . This below method will check for Any and convert it t0 NSNumber its easy to convert a NSNumber to Int, Float what ever data type you need.
Actual Code
//Convert Any To Number Object Removing Optional Key Word.
public func getNumber(number: Any) -> NSNumber{
guard let statusNumber:NSNumber = number as? NSNumber else {
guard let statString:String = number as? String else {
guard let statSubStr : Substring = number as? Substring else {
return 0
}
if let myInteger = Int(statSubStr) {
return NSNumber(value:myInteger)
}
else{
return 0
}
}
if let myInteger = Int(statString) {
return NSNumber(value:myInteger)
}
else if let myFloat = Float(statString) {
return NSNumber(value:myFloat)
}else {
return 0
}
}
return statusNumber }
Usage
if let hourVal = getNumber(number: hourStr) as? Int {
}
Passing String to check and convert to Double
Double(getNumber(number: dict["OUT"] ?? 0)
As of swift 3, I have to force my #%#! string & int with a "!" otherwise it just doesn't work.
For example:
let prefs = UserDefaults.standard
var counter: String!
counter = prefs.string(forKey:"counter")
print("counter: \(counter!)")
var counterInt = Int(counter!)
counterInt = counterInt! + 1
print("counterInt: \(counterInt!)")
OUTPUT:
counter: 1
counterInt: 2

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